Oscillations Important Questions

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Very Short Answer Questions [1 Mark Questions]

Ques. Define periodic motion with examples.

Ans. A motion of the body that repeats itself after regular intervals of time is called periodic motion.

Examples of periodic motion are

• Motion of the planet around the sun
• Motion of the bob of a simple pendulum

Ques. Define non-periodic motion with examples.

Ans. The motion of the body that does not repeat itself at a regular interval of time is called non-periodic motion.

Examples of non-period motion are

• Applying breaks in moving vehicles.
• Motion of tides in the sea.

Ques. What is meant by oscillatory motion?

Ans. Oscillatory motion is a special type of periodic motion. The motion of the body is said to be oscillatory motion if it moves to and fro about a fixed point after a regular interval of time.

Ques. Give an example of periodic motion which is not oscillatory.

Ans. The motion of the planet around the sun is a periodic motion but not an oscillatory motion.

Ques. Give an example of oscillatory motion which is also periodic.

Ans. To and fro motion of a pendulum is oscillatory motion, as well as periodic.

Ques. What is meant by the force constant of a spring?

Ans. The force constant, also known as the spring constant is a measure of the stiffness of a spring. It is defined as the force per unit deformation of the spring. The SI unit force constant is N/m.

Ques. What is meant by free oscillation?

Ans. A system is said to execute free oscillations if on being displaced from its mean position, it oscillates itself with a natural frequency.

Ques. What is meant by forced oscillation?

Ans. When a body is compelled to oscillate with a frequency other than its natural frequency, it is said to execute forced oscillations.

Ques. Define resonance.

Ans. It is a particular case of forced oscillation in which the frequency of the driving force is equal to the natural frequency of the system. This oscillation is called resonant oscillation and the phenomenon is called resonance.

Ques. What is meant by undamped oscillation?

Ans. If the frequency of the oscillating particle remains constant due to no loss of energy, then the oscillation is called undamped oscillation.

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Short Answer Questions [2 Marks Questions]

Ques. Define simple harmonic motion. Write its mathematical expression.

Ans. A particle is said to be executing simple harmonic motion, if it moves to and fro about a mean position under the action of restoring force which is directly proportional to its displacement from the mean position, and is always directed towards the mean position.

The mathematical expression for simple harmonic motion is given by

x = A sin ωt or 

x = x = A cos ωt

Ques. What is meant by angular simple harmonic motion?

Ans. When the oscillating particle moves on an arc of a circle and the restoring torque is directly proportional to the angular displacement from the mean position, it is called angular simple harmonic motion (angular SHM).

Ques. Write the formula of kinetic energy and potential energy in simple harmonic motion.

Ans. The kinetic energy of a particle executing SHM is given by

K.E.= (1/2)k(A² – x²)

The potential energy of a particle executing SHM is given by

U = (1/2)kx²

Where

• x is the displacement of the particle from the mean position
• A is the amplitude of the oscillation
• k is force constant.

Ques. State the laws of simple pendulum.

Ans. According to the laws of simple pendulum

• The time period of a simple pendulum is proportional to the square root of its length.
• The time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.
• The time period of a simple pendulum is not affected by its mass.
• The amplitude of a simple pendulum has no effect on its time period.

Ques. Write the equation of the time period if two springs of spring constant k₁ and k₂ are connected in series with a block of mass m.

Ans. If two springs of spring constant k₁ and k₂ are connected in series with a block of mass m, then the time period of the oscillations is given by

T = \(2 \pi \sqrt{\frac{k_1 + k_2 m}{k_1k_2}}\)

Ques. Write the equation of the time period if two springs of spring constant k₁ and k₂ are connected in parallel with a block of mass m.

Ans. If two springs of spring constant k₁ and k₂ are connected in parallel with a block of mass m, then the time period of the oscillations is given by

T = \(2 \pi \sqrt{\frac{m}{k_1+k_2}}\)

Ques. What is meant by maintained oscillation? Give an example.

Ans. It is possible to maintain the amplitude of the oscillation constant by supplying energy from an external source. We refer to these kinds of oscillations as maintained oscillations. 

The vibration of a tuning fork powered by an external power source or a battery is an example of maintained oscillations.

Also Read:

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Long Answer Questions [3 Marks Questions]

Ques. What are the characteristics of simple harmonic motion?

Ans. The following are the characteristics of simple harmonic motion:

• A restoring force must act on the body.
• The body must have acceleration in the opposite direction of displacement and the acceleration is directly proportional to the displacement.
• The system must have inertia (mass).
• It is a type of oscillatory motion and a particular case of periodic motion.
• .Energy of the system oscillates between kinetic and potential energy but the total energy remains constant, provided there is no loss in energy.

Ques. A particle executes SHM with amplitude A and time period T. When the displacement from the equilibrium position is half the amplitude, what fractions of the total energy are kinetic and potential?

Ans. The total energy of a particle executing SHM is given by

E = (1/2)kA² …(i)

Where

• A is the amplitude of the oscillation
• k is the force constant

The potential energy of a particle executing SHM is given by

U = (1/2)kx²

Where x is the displacement of the particle from the mean position

When the displacement from the equilibrium position is half the amplitude, i.e. x = ± A/2, then

U = (1/2)k(± A/2)²

⇒ U = (1/4) x (1/2)kA²

Using equation (i), we get

U = E/4 = 25% of total energy (E)

Now, kinetic energy, K.E. = E - U

⇒ K.E. = E - E/4

⇒ K.E. = 3E/4 = 75% of total energy(E)

Ques. If the length of a simple pendulum of a clock increases by 2%. How much loss or gain of second per day will take place?

Ans. The time period of a simple pendulum is given by

T = 2π√(L/g)

⇒ T ∝ L^1/2

Let T and L be the original time period and length, and ΔT and ΔL be the change in time period and change in length respectively, then

ΔT/T = (1/2)(ΔL/L)

Given, that the length of a simple pendulum of a clock increases by 2%

i.e. (ΔL/L) = 2% = 2/100

⇒ ΔT/T = (1/2) x (2/100) = 1/100 = 1%.

Therefore the time period will be increased by 1%.

Now, in one day, the time period in seconds, T = 86400 seconds

Therefore 1% of 86400 = 864 seconds

⇒ ΔT = 864 seconds

As length increases, the time period T will become large and the clock will go slow, so it will lose 864 seconds.

Ques. On an average human, the heartbeat is found to be 75 times in a minute. Calculate its beat frequency and period.

Ans. Given, the number of oscillations = 75 per minute

Frequency of the oscillation (ν) = (Number of oscillations)/(Time)

⇒ ν = 75/60 = 1.25 Hz

The time period of the oscillation is given by

T = 1/frequency(ν)

⇒ T = 1/1.25 = 0.8 seconds

That means, in one second it completes 1.25 oscillations, and in 0.8 seconds it completes 1 oscillation.

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Very Long Answer Questions [5 Marks Questions]

Ques. For the damped oscillator, the mass m of the block is 400 g, k = 45 N m^-1 and the damping constant is 80 g s^-1. Calculate

1. The time period of the oscillation
2. Time taken for its amplitude of vibrations to drop to half of its initial value and
3. The time taken for its mechanical energy to drop to half of its initial value.

Ans. Given

• Mass of the block, m = 400 g = 0.4 kg
• Restoring force constant, k = 45 N m^-1
• Damping constant, b = 80 g s-1 = 0.08 kg s^-1

1. The time period of the oscillation is given by

T = 2π√(m/k)

On substituting the values, we get

T = 2π√(0.4/45)

⇒ T = 0.6 second

Hence the time period of the oscillation is 0.6 seconds.

2. The displacement equation of the damped oscillation is given by

x(t) = A e^-(bt/2m) cos (ω’ + ϕ)

Where A e^-(bt/2m) is the amplitude at time t.

Now, at t = t’, A = A/2

⇒ A e^-(bt’/2m) = A/2

⇒ e^-(bt’/2m) = 1/2

⇒ bt’/2m = ln (2)

⇒ t’ = [ln (2) x 2m]/b

On substituting the values, we get

t’ = 6.93 seconds

3. Energy for the damped oscillator is given by

E = (1/2)kA²e^(-bt/m)

Initially, at t = 0, E = (1/2)kA² …..(i)

Finally, at t = t’, the energy of the oscillator reduces to half i.e. E = E/2

⇒ E/2 = (1/2)kA²e^(-bt’/m) ……(ii)

Dividing equation (ii) by equation (i), we get

(E/2)/E = [(1/2)kA²e^(-bt’/m)]/[(1/2)kA²]

⇒ 1/2 = e^(-bt’/m)

⇒ e^(bt’/m) = 2

⇒ (bt’/m) = ln (2)

⇒ t’ = [ln (2) x m]/b

On substituting the values, we get

t’ = 3.46 seconds.

Ques. Find the time period of oscillation of the block in the following arrangement. (Assume the pulley is ideal).

Ans. Let us assume under equilibrium condition extension in the spring is x₀.

Net upward tension on the spring is

T + T = 2T

Restoring force on spring due to displacement x₀ is given by

F = – kx₀

Upward tension on the spring provides restoring force i.e.

2T = – F

⇒ 2T = kx₀

⇒ T = kx₀/2

Now further the block of mass m is displaced by distance x. Then additional extension in the spring will be x/2.

So the net displacement of spring will be (x₀ +x/2).

Hence, restoring force will become, F’ = – k((x₀ +x/2)

Tension on the string will also changed to T’.

⇒ 2T’ = F’ = k((x₀ +x/2)

The net force acting on the block is given by

Fnet = mg - (k/2)((x₀ +x/2) = mg - kx₀/2 - kx/4

But mg = T = kx₀/2

⇒ Fnet = kx₀/2 - kx₀/2 - kx/4 = -kx/4

But Fnet = ma

⇒ ma = -kx/4

⇒ a = -(k/4m)x ….(i)

So effective force constant is k/4

Comparing equation (i) with a = - ω²x, we get

ω² = k/4m

⇒ ω = √(k/4m)

But ω = 2π/T

⇒ 2π/T = √(k/4m)

⇒ T = 2π√(4m/k)

Ques. A particle of mass 0.2 kg is executing a simple harmonic motion of amplitude 0.2 m. When it passes through the mean position, its kinetic energy is 64 x 10-3 J. Obtain the equation of motion of this particle if the initial phase of oscillation is π/4.

Ans. Given

• The amplitude of the oscillation, A = 0.2 m
• Initial phase, ϕ = π/4
• The mass of the particle, m = 0.2 kg
• The kinetic energy at the mean position, K.E. = 64 x 10^-3 J

The kinetic energy at distance x from the mean position is given by

K.E. = (1/2)mω² (A² - x²)

At the mean position (x = 0), K.E. = 64 x 10^-3 J

⇒ 64 x 10^-3 = (1/2)mω² (A² - 0)

⇒ 64 x 10^-3 = (1/2)mω²A²

On substituting the values of m and A, we get

64 x 10^-3 = (1/2) x 0.2 x ω² x 0.2²

⇒ ω² = 16

⇒ ω = 4 rad/s

Hence the equation of motion at the mean position is given by

x = A sin(ωt + ϕ)

⇒ x = 0.2 sin(4t + π/4)

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